Question: Multiply the following complex numbers: $({4+3i}) \cdot ({4+2i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({4+3i}) \cdot ({4+2i}) = $ $ ({4} \cdot {4}) + ({4} \cdot {2}i) + ({3}i \cdot {4}) + ({3}i \cdot {2}i) $ Then simplify the terms: $ (16) + (8i) + (12i) + (6 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 16 + (8 + 12)i + 6i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 16 + (8 + 12)i - 6 $ The result is simplified: $ (16 - 6) + (20i) = 10+20i $